Re: Equation of the tangent to a perpendicular lin
Originally Posted by mgmanoj
After differentiating the equation I got dy = 2x-2y-7
the slope of the tangent perpendicuar to 6x + 5y-4 = 0 is 45/6 or 5/6 - I got confused here.
The equation 6x+ 5y- 4= 0 is the same as y= -(6/5)x+ 4/6 and has slope -6/5. The "negative reciprocal" of that is 6/5 and is the slope of the perpendicular line. I don't know where you got "45/6"! Did you mean "4/6"? Remember that the slope of the line, the number multiplying x, determines the direction, not the "constant" term.